poisson systems and perfect sampling -...

MARKOV MODELS PERFECT SAMPLING DISCRETE TIME EVENT SIMULATION CASE STUDIES Ψ3 -SOFTWARE DEMONSTRATION S

Poisson Systems and Perfect Sampling

[email protected]

Laboratoire d’Informatique de Grenoble, INRIA Team Polaris

École d’été en Recherche OpérationnelleOptimisation et décision en milieu incertain

June 4-6, 2016

Work partially supported by ANR Marmote

1 / 105Poisson Systems and Perfect Sampling


MARKOVIAN WORK

An example of statistical investigation in the text of"Eugene Onegin" illustrating coupling of "tests" inchains.(1913) In Proceedings of Academic Scientific St.Petersburg, VI, pages 153-162.

1856-1922



GRAPHS AND PATHS

i

1

2

3

4 5

a

bc

d

e f

g

hRandom Walks

Path in a graph :Xn n-th visited nodepath : i0, i1, · · · , innormalized weight : arc (i, j) −→ pi,j

concatenation : . −→ ×P(i0, i1, · · · , in) = pi0,i1 pi1,i2 · · · pin−1,in

disjoint union : ∪ −→ +P(i0 ❀ in) =

∑

i1,··· ,in−1pi0,i1 pi1,i2 · · · pin−1,in

automaton : state/transitions randomized (language)



MODELING AND ANALYSIS OF COMPUTER SYSTEMS

Complex system

output

Environment

Input of the system

System

System

Basic model assumptions

System :- automaton (discrete state space)- discrete or continuous timeEnvironment : non deterministic- time homogeneous- stochastically regular

Problem

Understand “typical” states- steady-state estimation- ergodic simulation- state space exploring techniques



QUEUING NETWORKS WITH FINITE CAPACITY

Network model

Finite set of resources :◮ servers

◮ waiting rooms

Routing strategies :◮ state dependent

◮ overflow strategy

◮ blocking strategy

◮ ...

Average performance :◮ load of the system

◮ response time

◮ loss rate

◮ ...

Markov model

1−p

λ

µ

ν

C1

C2 Rejection

Blocking

p

Poisson arrival, exponential services distribution, probabilistic routing⇒ continuous time Markov chain

C1

0

1

2

C2−1

C2

Queue 2

0 2 Queue 11 C1−1

Problem

Computation of steady state distribution⇒ state-space explosion



INTERCONNEXION NETWORKS

Delta network

8

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

0

1

2

3

4

5

6

7

10

9

Input ratesService ratesHomogeneous routingOverflow strategy

Problem

Loss probability at each levelAnalysis of hot spot...



CALL CENTERS

Multilevel Erlang model

Traffic 2

Overflow traffic 2Overflow traffic 1

Type I servers Type II servers

Type III servers

Traffic 1

Types of requestsInput ratesDifferent service ratesOverflow strategy

Problem

Optimization of resourcesQuality of service (waiting time, rejection probability,...)...



RESOURCE BROKER

Grid model

Q9

Resource Broker

λ

µ2

µ1

µ3

µ4

µ5

µ6

µ7

µ8

µ9

Overflow

Q2

Q3

Q1

Q10

Q11

Q12

Q4

Q6

Q5

Q7

Q8

Input ratesAllocation strategyState dependent allocationIndex based routing : destinationminimize a criteria

Problem

Optimization of throughput, response time,...Comparison of policies, analysis of heuristics...



FORMAL DEFINITION

Let {Xn}n∈Na random sequence of variables in a discrete state-space X

{Xn}n∈Nis a Markov chain with initial law π(0) iff

◮ X0 ∼ π(0) and

◮ for all n ∈ N and for all (j, i, in−1, · · · , i0) ∈ X n+2

P(Xn+1 = j|Xn = i, Xn−1 = in−1, · · · , X0 = i0) = P(Xn+1 = j|Xn = i).

{Xn}n∈Nis a homogeneous Markov chain iff

◮ for all n ∈ N and for all (j, i) ∈ X 2

P(Xn+1 = j|Xn = i) = P(X1 = j|X0 = i)def= pi,j.

(invariance during time of probability transition)



ALGEBRAIC REPRESENTATION

P = ((pi,j)) is the transition matrix of the chain

◮ P is a stochastic matrixpi,j > 0;

∑

j

pi,j = 1.

Linear recurrence equation πi(n) = P(Xn = i)

πn = πn−1P.

◮ Equation of Chapman-Kolmogorov (homogeneous) : Pn = ((p(n)i,j

))

p(n)i,j

= P(Xn = j|X0 = i); Pn+m= Pn

.Pm;

P(Xn+m = j|X0 = i) =∑

k

P(Xn+m = j|Xm = k)P(Xm = k|X0 = i);

=∑

k

P(Xn = j|X0 = k)P(Xm = k|X0 = i).

Interpretation : decomposition of the set of paths with length n + m from i to j.



EVENT MODELLING

Multidimensional state space : X = X1 × · · · × XK with Xi = {0, · · · ,Ci}.Event e :❀ transition function Φ(., e) ; (skip rule)❀ Poisson process λe

States

Events

e1

e2

e3

e4

Time

1781-1840



EVENT MODELLING : UNIFORMIZATION

Λ =∑

e

λe and P(event e) =λe

Λ;

Trajectory : {en}n∈Zi.i.d. sequence.

⇒ Homogeneous Discrete Time Markov Chain [Bremaud 99]

Xn+1 = Φ(Xn, en+1).

Generation among a small finite space E :O(1)



EVENTS AND POISSON SYSTEMS

0

Queue

States

2

M/M/1 capacity C = 2

1

Φ(x, d) = max(x − 1, 0)Φ(x, a) = min(x + 1,C)

non effective eventsState

1

2

0

arrivals

departures

time

⇒monotone events



PROBLEMS

Finite horizon

- Estimation of π(n)- Estimation of stopping times

τA = inf{n > 0; Xn ∈ A}

- · · ·

Infinite horizon

- Convergence properties- Estimation of the asymptotics- Estimation speed of convergence- · · ·



ASYMPTOTIC BEHAVIOR : LAW CONVERGENCE

Let {Xn}n∈Na homogeneous, irreducible and aperiodic Markov chain taking values in

a discrete state X then◮ The following limits exist (and do not depend on i)

limn→+∞

P(Xn = j|X0 = i) = πj;

◮ π is the unique probability vector invariant by P

πP = π;

◮ The convergence is rapid (geometric) ; there is C > 0 and 0 < α < 1 such that

||P(Xn = j|X0 = i) − πj|| 6 C.αn.

DenoteXn

L−→ X∞;

with X∞ having law ππ is the steady-state probability associated to the chain



INTERPRETATION

Equilibrium equation

j,j

j

i1

i2

i3

i4

k1

k2

k3

p

p

p

p

p

p

p

p

i1,j

i2,j

i3,j

i4,j

j,k1

j,k2

j,k3

Probability to enter j =probability to exit jbalance equation

∑

i6=j

πipi,j =∑

k 6=j

πjpj,k = πj

∑

k 6=j

pj,k = πj(1− pj,j)

πdef= steady-state.

If π0 = π the process is stationary (πn = π)



ERGODIC THEOREM

Let {Xn}n∈Na homogeneous aperiodic and irreducible Markov chain on X with

steady-state probability π then- for all function f satisfying Eπ |f | < +∞

1N

N∑

n=1

f (Xn)P−p.s.−→ Eπ f .

generalization of the strong law of large numbers- If Eπ f = 0 then there exist σ such that

1

σ√

N

N∑

n=1

f (Xn)L−→ N (0, 1).

generalization of the central limit theorem



FUNDAMENTAL QUESTION

Given a function f (cost, reward, performance,...) estimate

Eπf

and provide the quality of this estimation.



SOLVING METHODS

Solving π = πP

◮ Analytical/approximation methods

◮ Formal methods N 6 80Maple, Sage,...

◮ Direct numerical methods N 6 5000Mathematica, Scilab,...

◮ Iterative methods with preconditioning N 6 500, 000Marca,...

◮ Adapted methods (structured Markov chains) N 6 10, 000, 000PEPS,...

◮ Monte-Carlo simulation N > 108

Postprocessing of the stationary distribution

Computation of rewards (expected stationary functions)Utilization, response time,...



ERGODIC SAMPLING(1)

Ergodic sampling algorithm

Representation : transition fonction

Xn+1 = Φ(Xn, en+1).

x ← x0{choice of the initial state at time =0}n = 0 ;repeat

n ← n + 1 ;e ← Random_event() ;x ← Φ(x, e) ;Store x{computation of the next state Xn+1}

until some empirical criteriareturn the trajectory

Problem : Stopping criteria



ERGODIC SAMPLING(2)

Start-up

Convergence to stationary behavior

limn→+∞

P(Xn = x) = πx.

Warm-up period : Avoid initial state dependenceEstimation error :

||P(Xn = x)− πx|| 6 Cλn2 .

λ2 second greatest eigenvalue of the transition matrix- bounds on C and λ2 (spectral gap)- cut-off phenomena

λ2 and C non reachable in practice(complexity equivalent to the computation of π)some known results (Birth and Death processes)



ERGODIC SAMPLING(3)

Estimation quality

Ergodic theorem :

limn→+∞

1n

n∑

i=1

f (Xi) = Eπ f .

Length of the sampling : Error control (CLT theorem)

Complexity

Complexity of the transition function evaluation (computation of Φ(x, .))Related to the stabilization period + Estimation time



ERGODIC SAMPLING(4)

Typical trajectory

States

0 time

Warm−up period Estimation period



REPLICATION METHOD

Typical trajectory

States

0 timereplication periods

Sample of independent statesDrawback : length of the replication period (dependence from initial state)



REGENERATION METHOD

Typical trajectory

States

0 time

start−up period

regeneration period

R1 R2 R3 ....

Sample of independent trajectoriesDrawback : length of the regeneration period (choice of the regenerative state)



STOCHASTIC RECURSIVE SEQUENCES

Description [Borovkov et al]

◮ Discrete state space X (usually lattice, product of intervals,...)

◮ Innovation state space, and an innovation process

◮ Dynamic of the system : transition function

Φ : X × E −→ X(x, ξ) 7−→ y

◮ Trajectory given by x0 and {ξn} an innovation process

X0 = x0; Xn+1 = Φ(Xn, ξn)

Discrete event systems

◮ state space : usually lattice, product of intervals,...

◮ Innovations : usually a set of events E

◮ Independent innovation process : Poisson systems (uniformization)



MARKOVIAN MODELLING

Theorem (Markov process)

If {ξn} is a sequence of iid random variables , the process {Xn} is a homogeneous discrete timeMarkov chain.

Random Iterated system of functions

The trajectory Xn is the successive application of random functions taken in the set{Φ(., ξ), ξ ∈ E} according a probability measure on E[Diaconis and Friedman 98]



COUPLING INEQUALITY

Typical trajectory

Coupling time

Stationary version

τ0 time

States

After τ the two processes are not distinguishable, then stationaryScheme used to prove Markov convergence (coupling inequality)

|P(Xn ∈ A)− πA| 6 P(τ > n)



FORWARD SAMPLING : AVOID INITIAL STATE DEPENDENCE

Forward coupling

Steady-state ?

f3f4f6f7f3f1Time

State

Example

1

1 − p p

Always couple in the blue stateDoes not guarantee the steady state !



PERFECT SAMPLING : BACKWARD IDEA

Set dynamic

In what state could I be at time n = 0 ?

X0 ∈ X = Z0

∈ Φ(X , e−1) = Z1

∈ Φ(Φ(X , e−2), e−1) = Z2

......

∈ Φ(Φ(· · ·Φ(X , e−n), · · · ), e−2), e−1) = Zn



PERFECT SAMPLING : BACKWARD IDEA

All the trajectories

Time

0

−i

−j

−τ ∗

Stationary Process

X

X

X

X

Zi

Zj

Z−τ∗ = {X0}

Z0 = X

collapse



PERFECT SAMPLING : CONVERGENCE THEOREM

Theorem

Provided some condition on the events the sequence of sets

{Zn}n∈N

is decreasing to a single state, stationary distributed.

τ∗ = inf{n ∈ N; Card(Zn) = 1}.backward coupling time

The set of possible states at time 0 is decreasing with regards to n



PERFECT SAMPLING : COUPLING CONDITION

Theorem

Suppose that the set of events is finite. Then the two conditions are equivalent :◮ τ∗ < +∞ almost surely ;

◮ There exist a finite sequence of events with positive probability S = {e1, · · · , eM} such that

|Φ(X , e1→M)| = 1.

The sequence S is called a synchronizing pattern(synchronizing word, renovating event,...)



PERFECT SAMPLING : COUPLING CONDITION (PROOF)

Proof

⇒ If τ∗ < +∞ almost surely there is a trajectory that couples in a finite time. This finitetrajectory is a synchronizing pattern.

⇐ Suppose there is a synchronizing pattern with length M. Because the sequence of events is iid,it occurs almost surely on every trajectory. Applying Borel-Cantelli lemma gives the result.

The forward and backward coupling time have the same distributionτ∗ has an exponentially dominated distribution tail

P(τ∗ > M.n) 6 (1− P(e1→M))n.

Practically efficient



PERFECT SAMPLING : CONVERGENCE THEOREM (PROOF 1)

Proof based on the shift property

First, because τ < +∞ and the ergodicity of the chain there exists N0 s.t.

|P(Φ(X , e1→n) = {x})− πx| 6 ǫ.

But the sequence of events is iid (stationary) then

P(Φ(X , e1→n) = {x}) = P(Φ(X , e−n+1→0) = {x})

τ∗ < +∞ then there exists N1 such that P(τ∗ > N1) 6 ǫ ; then

P(Φ(X , e−n+1→0) = {x})

= P(Φ(X , e−n+1→0) = {x}, τ∗ < N1) + P(Φ(X , e−n+1→0) = {x}, τ∗ > N1),

= P(Φ(X , e−τ∗→0) = {x}, τ∗ < N1) + ǫ′,

= P(X0 = x, τ∗ < N1) + ǫ′ = P(X0 = x) + ǫ′′.

By the dominated convergence theorem the result follows



PERFECT SAMPLING : CONVERGENCE THEOREM (PROOF 2)

Proof based on the coupling property [Haggstrom]

Consider N such that P(τ∗ > N) 6 ǫthen consider a process {Xn}with the same events e−N+1→0 but with X−N+1

generated according π. The process {Xn} is stationary.On the event (τ∗ < N) we have X0 = X0 and

P(X0 6= X0) 6 P(τ∗ > N) 6 ǫ (coupling inequality).

FinallyP(X0 = x)− πx = P(X0 = x)− P(X0 = x) 6 P(X0 6= X0) 6 ǫ;

πx − P(X0 = x) = P(X0 = x)− P(X0 = x) 6 P(X0 6= X0) 6 ǫ;

and the result follows.



PERFECT SAMPLING : ALGORITHM

Backward algorithm


Xn+1 = Φ(Xn, en+1).

for all x ∈ X doy(x) ← x

end forrepeat

e ← Random_event() ;for all x ∈ X do

z(x) ← y(Φ(x, e)) ;end fory ← z

until All y(x) are equalreturn y(x)

Convergence : If the algorithm stops, the returnedvalue is steady state distributedCoupling time : τ < +∞, properties of Φ

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

τ∗

Mean time complexity

cΦ mean computation cost of Φ(x, e)

C 6 Card(X ).Eτ.cΦ.



PERFECT REWARD SAMPLING

Backward reward


Xn+1 = Φ(Xn, en+1).

Arbitrary reward function

for all x ∈ X doy(x) ← x

end forrepeat

e ← Random_event() ;for all x ∈ X do

y(x) ← y(Φ(x, e)) ;end for

until All Reward(y(x)) are equalreturn Reward(y(x))

Convergence : If the algorithm stops, the returnedvalue is steady state reward distributedCoupling time : τr 6 τ < +∞

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

1

0

Cost


CReward 6 Card(X ).Eτ.cΦDepends on the reward function.



INVERSE OF PDF

P(X 6 x)

0

1

1 2 3 K − 1 K

Cumulative distribution function

x

p1

p2

p3

· · ·

Generation

Divide [0, 1[ in intervals with length pk

Find the interval in which Random fallsReturns the index of the intervalComputation cost :O(EX) stepsMemory cost :O(1)

Inverse function algorithm

s=0 ; k=0 ;u=random()while u >s do

k=k+1s=s+pk

end whilereturn k



SEARCHING OPTIMIZATION

Optimization methods

◮ pre-compute the pdf in a table

◮ rank objects by decreasing probability

◮ use a dichotomy algorithm

◮ use a tree searching algorithm (optimality = Huffmann coding tree)

Comments

- Depends on the usage of the generator (repeated use or not)- pre-computation usuallyO(K) could be huge-



GENERATION : VISUAL REPRESENTATION

[0, 1] partitionning

10312

112

612

212

712

112

412

312

712

312

1112

f1 f3 f4 f5 f6 f7 f8f2

4 3

21

Random iterated system of functions

Function f1 f2 f3 f4 f5 f6 f7 f8Probability 1

121

121

121

121

124

122

121

12

Stochastic matrix P =⇒ simulation algorithm = RIFS



THE COUPLING PROBLEM

τ estimation

Eτ = 2

0 1

Couples with probability 12

0 1

Never couples

τ = ∞

10 1

Couples with probability 1

τ = 1

0

12

12

12

12



GENERAL PROBLEM

Objective

Given a stochastic matrix P = ((pi,j)) build a system of function (fθ, θ ∈ Θ) and a probabilitydistribution (pθ, θ ∈ Θ) such that :

1 the RIFS implements the transition matrix P,

2 ensures coupling in finite time

3 achieve the “best” mean coupling time : tradeoff between- choice of the transition function according to ((pθ)),- computation of the transition

Remarks

Usual method|Θ| = number of non-negative elements of P = O(n2

)

choice inO(log n)



NON SPARSE MATRICES

Rearranging the system

0 1

"Synchronizing" transformation

0 1

312

112

412

712

312

312

1112

112

612

212

712

f1 f3 f4 f5 f6 f7 f8f2

4 3

21



NON SPARSE MATRICES

Convergence rate

When at least one column is non-negative⇒ one step coupling.The RIFS ensures coupling and the coupling time τ is upper bounded by a geometricdistribution with rate

∑

j

mini

pi,j

number of transition functions : could be more than the number of non-negativeelementsat most n2



ALIASING TECHNIQUE

Initialization

K objectslist L=∅,U=∅ ;for k=1 ; k6 K ; k++ do

P[k]=pk

if P[k] > 1K

thenU=U+{k} ;

elseL=L+{k} ;

end ifend for

Alias and threshold tables

while L 6= ∅ doExtract k ∈ LExtract i ∈ US[k]=P[k]A[k]=iP[i] = P[i] - ( 1

K-P[k])

if P[i] > 1K

thenU=U+{i} ;

elseL=L+{i} ;

end ifend while

Combine uniform and alias value when rejection



ALIASING TECHNIQUE : GENERATION

1/8



ALIASING TECHNIQUE : GENERATION

Generation

k=alea(K)if Random . 1

K6 S[k] then

return kelse

return A[k]end if

Complexity

Computation time :-O(K) for pre-computation-O(1) for generationMemory :- thresholdO(K) (real numbers as probability)- aliasO(K) (integers indexes in a tables)



SPARSE MATRICES

Rearranging the system

10

Aliasing transformation

10

612

1112

312

312

712

412

112

312

712

212

112

f1 f3 f4 f5 f6 f7 f8f2

4 3

21

Complexity

M = maximum out degree of statespθ uniform on {1, · · · , M}, threshold comparisonO(1) to compute one transitionCombination with “Synchronizing” techniques



UNIFORM-BINARY DECOMPOSITION

Uniform superposition

1

1

1

1

1

1

34

2

A4

A4

13

1

23

131

22

1 2

34

A3

A3

1

1

1

1

4

A2

A2

21

3

13

23

13

1

1

23

4 3

21

A1

A1

0 1

Aliasing transformation

312

412

112

712

212

612

112

312

1112

312

712

1 2

34

Decomposition

P =1

M

M∑

i=1Pi, Pi : stochastic matrix with at most 2 non negative elements per row



COUPLING PROPERTY

Exchange of columns or thresholds give an equivalent representative

representation

��

��

��

��

��

��

��

��

��

��

��

��

Equivalent

Spanning tree

Irreducibility =⇒ there is a spanning tree going to a single state where coupling occurs.

P(τ∗ < +∞) = 1.

τ is geometrically bounded, soτ∗ and τ∗C .



Ψ SOFTWARE



EXAMPLE

Random transition coefficients :

Number of states 10 100 500 1000 3000Mean coupling time 3.1 4.5 5.3 5.7 6.1Mean execution time µs 3 17 170 360 1100

Pentium III 700MHz and 256Mb memory. Sample size 10000.Remarks :- very small coupling time- Coefficients : same order of magnitude, aliasing enforces coupling

Comparison with birth and death process :

Number of states 10 100 500 1000 3000Mean coupling time 41 557 2850 5680 17000Mean execution time µs 28 1800 88177 366000 3.5s

Remarks :- large coupling time- sparse matrix, large graph diameter



OVERFLOW MODEL

Model

K

Overflow

rejection

λ µ1

µ2

µ3

µ

Coupling time distribution

Coupling time (iterations)

Den

sity

Sample size : 10000

0

0.05

0.1

0.15

0.2

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000

Parameters

K servers,priority on overflowsinput rate λ,different service ratestate (x1, · · · , xK), xi ∈ {0, 1}, size∼ 130000low diameternon product-form structure,

Statistics

Parameter Valueminimum 113maximum 1794median 465mean 498Std 180

exponential tail, low mean value



OVERFLOW MODEL (2)

Marginal distribution

Index of the server

Util

izat

ion

of s

erve

r

Sample size 10000

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 K

Marginal probability estimation

P(Xi = 1)

Occupied servers

Sample size = 10000

Number of occupied servers

Pro

babi

lity

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Marginal distribution of the occupiedservers



OVERFLOW MODEL (3)

Reward coupling

mean value

Cou

plin

g tim

e (it

erat

ions

)

Marginal law (server number)

Sample size 10000 Maximum

Quartile 3Median

Minimum

Quartile 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0

500

1000

1500

2000

2500

loadoc

cupa

tion

stat

e

Reward coupling time- gain 20% for the first marginals- utilization : best reduction



MONOTONICITY AND PERFECT SAMPLING : IDEA

(X ,≺) partially ordered set (lattice)

Typically componentwise ordering on products of intervals

min = (0, · · · , 0) and Max = (C1, · · · ,Cn).

An event e is monotone if Φ(., e) is monotone on XIf all events are monotone then

X0 ∈ Zn ⊂ [Φ(min, e−n→0),Φ(Max, e−n→0)]

⇒ 2 trajectories



THE DOUBLING SCHEME

Complexity

◮ Need to store the backward sequence of events

◮ Consider 2 trajectories issued from {min, Max} at time −n and test if couplingOne step backward ⇒

2.(1 + 2 + · · · + τ∗) = τ

∗(τ∗+ 1) = O(τ

∗2)

calls to the transition function.

◮ Consider 2 trajectories issued from {min, Max} at time −2k and test if couplingDoubling step backward ⇒

2.(1 + 2 + · · · + 2k) = 2k+2

− 2

calls to the transition function, with k such that 2k−1 < τ∗ 6 2k ,Number of calls : O(τ∗)



MONOTONICITY AND PERFECT SAMPLING

Monotone PS

Doubling schemen=1 ;R[1]=Random_event ;repeat

n=2.n ;y(min) ← miny(Max) ← Maxfor i=n downto n/2+1 do

R[i]=Random_event ;end forfor i=n downto 1 do

y(min) ← Φ(y(min), R[i])y(Max) ← Φ(y(Max), R[i])

end foruntil y(min) = y(Max)return y(min)

Trajectories

State

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

: minimum

Generated

0


Cm 6 2.(2.Eτ).cΦ. Reduction factor : 4Card(X )

.



INDEX ROUTING IN QUEUING NETWORKS

Index functions for event e

For queue i Iei : {0, · · · ,Ci} −→ O (totally ordered set).

Property : ∀xi, xj Iei (xi) 6= Ie

j (xj).

ex : inverse of a priority,...

Routing algorithm :

if xorigin >0 then{ a client is available in the origin queue}xorigin = xorigin − 1 ; { the client is removed from the origin queue}j = argmini Ie

i (xi) ; { computation of the destination}if j 6= -1 then

xj = xj+1 ; { arrival of the client in queue j }{ in the other case, the client goes out of the network}

end ifend if



MONOTONICITY OF INDEX ROUTING POLICIES

Proposition

If all index functions Iei are monotone then event e is monotone.

Proof :

Let x ≺ y two states and let be an index routing event. Let i be the origin queue for theevent.

jx = argminjIej (xj) and jy = argminjI

ej (yj)

Case 1 xi = yi = 0 nothing happens and Φ(x, e) = x ≺ y = Φ(y, e)

Case 2 xi = 0, yi > 0 then Φ(x, e) = x ≺ y− ei + ejy = Φ(y, e)

Case 3 xi > 0, yi > 0 then

Iejx(xjx ) < Ie

jy(xjy ) 6 Ie

jy(yjy ) < Ie

jx(yjx );

then xjx < yjx and

Φ(x, e) = x− ei + ejx 6 y− ei 6 y− ei + ejy = Φ(y, e)



MONOTONICITY OF ROUTING

Examples [Glasserman and Yao]

All of these events could be expressed as index based routing policies :- external arrival with overflow and rejection- routing with overflow and rejection or blocking- routing to the shortest available queue- routing to the shortest mean available response time- general index policies [Palmer-Mitrani]- rerouting inside queues...



MONOTONICITY OF ROUTING : EXAMPLES

Stateless routing

Overflow routing

Iej (xj) =

{

prio(j) if xj < Cj;+∞ elsewhere

Ie−1 = max

jCj.

Routing with blocking

Iej (xj) =

{

prio(j) if xj < Cj;+∞ elsewhere

Iei = max

jCj.

State dependent routing

Join the shortest queue

Iej (xj) =

{

xj if xj < Cj;+∞ elsewhere ;

Ie−1 = max

jCj.

Join the shortest response time

Iej (xj) =

{

xj+1µj

if xj < Cj;

+∞ elsewhere ;

Ie−1 = max

iCi.



COUPLING EXPERIMENT

Feed-forward queuing model

5

C

C

C

C0

1

2

3λ

λ

λ

λ

λλ

01

2

3

4

Estimation of Eτ :

0

50

100

150

200

250

300

350

400

τ

0 1 2 3 4 λ



MAIN RESULT

Bound on coupling time

Eτ 6

K∑

i=1

Λ

Λi

Ci + C2i

2,

- Λ : global event rate in the network,

- Λi the rate of events affecting Qi

- Ci is the capacity of Queue i.

Sketch of the proof

- Explicit computation for the M/M/1/C

- Computable bounds for the M/M/1/C

- Bound with isolated queues



EXPLICIT COMPUTATION FOR THE M/M/1/C

Eτ b = Emin(h0→C, hC→0)Absorbing time in a finite Markov chain ; p = λ

λ+µ= 1− q

1,C

1,C−10,C−2

1,C−2 2,C−1

2,C

3,C

C−2,C−1 C−1,C

C,C0,0

0,1 1,2

0,C

0,C−1

0,C−3

p

p

p

p

p

pp p p p

ppp

p p

pq

q

q

q

q

q q q q q

qqq

q q

qLevel 3

Level 4

Level 5

Level C+1

Level C+2

Level 2

Explicit recurrence equations

Case λ = µ Eτ b = C+C2

2 .



COMPUTABLE BOUNDS FOR M/M/1/C

If the stationary distribution is concentrated on 0 (λ < µ),

Eτ b6 Eh0→C is an accurate bound.

Theorem

The mean coupling time Eτ b of a M/M/1/C queue with arrival rate λ and service rate µ isbounded using p = λ/(λ+ µ) = 1− q.

Critical bound : ∀p ∈ [0, 1], Eτ b 6C2+C

2 .

Heavy traffic Bound : if p > 12 , Eτ b 6

Cp−q−

q(1−(

qp

)C)

(p−q)2 .

Light traffic bound : if p < 12 , Eτ b 6

Cq−p−

p(1−(

pq

)C)

(q−p)2 .



COMPUTABLE BOUNDS FOR M/M/1/C

Example with C = 10

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8 1

Eτ b

p

heavy trafficLight trafficbound

C+C2

2

C + C2

bound



EXAMPLE FOR TANDEM QUEUES

Coupling of Queue 0

Time

0

X00 = 55

4

2

1

3 = C1

6 = C0

0

−τ b0

Coupling of queue 1 conditionned by state of queue 0

4

3 = C1

1

0

−τ b1 (s0 = 2) 0

Time

X11 = 2

X10 = 3

5 = X00

6

2

X11 = 2

5

4

2

1

3 = C1

6 = C0

0

−τ b0 − τ b

1 (s0 = 5)

X00 = 5

τ b1 (s0 = 5) 0

Time

X10 = 3

Then τ b 6st∞τ b

1 + τ b0 , normalized



BOUND WITH ISOLATED QUEUES

Theorem

In an acyclic stable network of K M/M/1/Ci queues with Bernoulli routing and loss ifoverflow, the coupling time from the past satisfies in expectation,

E[τ b] 6

K−1∑

i=0

Λ

ℓi + µi

Ci

qi − pi−

pi(1−(

piqi

)Ci)

(qi − pi)2

6

K−1∑

i=0

Λ

ℓi + µi(Ci + C2

i ).



CONJECTURE FOR GENERAL NETWORKS

0

700

800

0 0.5 1 1.5 2 2.5 3 3.5 4

500

400

300

200

100

600

λ5

Eτ b

B1 (proven)

B1 ∧ B2 ∧ B3

B3 (conjecture)

B2 (conjecture)

Extension to cyclic networks,Generalization to several types of eventsApplication : Grid and call centers



PERFECT SAMPLING PRINCIPLE


Time

0

−i

−j

−τ ∗

Stationary Process

X

X

X

X

Zi

Zj

Z−τ∗ = {X0}

Z0 = X

collapse

Synchronizing pattern =⇒ finite backward scheme τ∗ <∞[NSMC 2003, LAA 2004]



MONOTONE PERFECT SAMPLING

−(n + 1)

X

X

time−n

m′M

m

M′

same convergence conditioncomplexity inO(Eτ∗)⇒ polynomial in model dimension

[NSMC 2006,QEST 2008]



ENVELOPES PERFECT SAMPLING

time

X

X

−(n + 1)

−n

m′

M′

Synchronizing pattern for envelopescomplexity unknown but practically efficient



ENVELOPES AND SPLITTING PERFECT SAMPLING

Exhaustive state sim

ulation

−i

−j

−τ ∗

Stationary Process

X

X

X

X

Splitting point

Envelope algorithm

Time

0

Guarantees the convergencecomplexity unknown but practically more efficient

[VALUETOOLS 2008]



PRIORITY SERVERS

Erlang model

Output

ArrivalsServers

Overflow

on next free server

Rejection if all servers

are buzy

X = {0, 1}3

E = {e0, e1, e2, e3}

Card(X ) = 2K

Events

Event type Rate Origin Destination listArrival λ −1 Q1 ; Q2 ; Q3 ; −1Departure µ1 Q1 −1Departure µ2 Q2 −1Departure µ3 Q3 −1

Results

◮ Validation χ2 test

◮ K = 30 µi decreasing

◮ Saturation probability 0.0579 ± 4.710−4

◮ Simulation time 0.4ms

◮ τ = 577



PRIORITY SERVERS

Erlang model

Output

ArrivalsServers

Overflow

on next free server

Rejection if all servers

are buzy

X = {0, 1}40

µ1 = 1,µ2 = 0.8,µ3 = 0.5Sample size 5.106

Card(X ) = 2K

Saturation probability

Prob

0.1

0.2

0.3

0.4

0.5

0.6

0 10 20 30 40 50 60 70 λ 0

Proba

1e−06

1e−04

0.001

8 8.5 9 9.5 10 10.5 11 11.5 λ

1e−05

Coupling time

s

Reward coupling time

Coupling time

0

200

250

300

350

0 10 20 30 40 50 60 70 λ

100

50

µ

400

150



LINE OF SERVERS

Tandem queues

reject

µ2C2

C3 µ3

C4 µ4

C5 µ5

C1 µ1

λ

X = {0, · · · , 100}5

E = {e0, · · · , e5}

Card(X ) = CK

Events

Event type Rate Origin Dest. listArrival λ −1 Q1 ; −1Routing/block µ1 Q1 Q2 ; Q1Routing/block µ2 Q2 Q3 ; Q2· · · · · · · · · · · ·Departure µ5 Q5 −1

Results

◮ C = 100 λ = 0.9 ; µ = 1 p = 12

◮ Blocking probability b1 = 0.34, b2 = 0.02 b3 = 0.02, b4,= 0.02.

◮ Simulation time < 1ms



MULTISTAGE NETWORK

Delta network8

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

0

1

2

3

4

5

6

7

10

9

X = {0, · · · , 100}32

E = {e0, · · · , e64}

Card(X ) = CK

Events

Event type Rate Origin Dest. listArrival λ −1 Qi ; −1Routing/rejection 1

2µ Qi Qj ; −1· · · · · · · · · · · ·Departure µ Qk −1

Results

◮ C = 100 λ = 0.9 ; µ = 1

◮ Loss rate

◮ Simulation time 135ms

◮ τ ≃ 400000



MULTISTAGE NETWORK

Delta network8

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

0

1

2

3

4

5

6

7

10

9

X = {0, · · · , 100}32

E = {e0, · · · , e64}

Card(X ) = CK

Sample size 100000

Queue length and saturation proba at level 3

λ

1

2

3

4

5

6

7

E(N_34)

0 0.2 0.4 0.6 0.8 0 λ

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78 0

Coupling time

sGlobal coupling

Reward at least 1 queue saturated (3rd level)

Reward queue 31 saturated

0

10000

12000

14000

16000

18000

0 0.2 0.4 0.6 0.8

6000

4000

2000

λ

µ

8000



RESOURCE BROKER

Grid model

Q9

Resource Broker

λ

µ2

µ1

µ3

µ4

µ5

µ6

µ7

µ8

µ9

Overflow

Q2

Q3

Q1

Q10

Q11

Q12

Q4

Q6

Q5

Q7

Q8

Input ratesAllocation strategyState dependent allocationIndex based routing : destinationminimize a criteria

Problem

Optimization of throughput, response time,...Comparison of policies, analysis of heuristics...



ROUTING CUSTOMERS IN PARALLEL QUEUES

The problem :

◮ Find a routing policy maximizing the expected (discounted) throughput of the system.

◮ Several variations on this problem depend on the information available to the controller :current size of all queues (and size of the arriving batch).

The applications :

◮ improve batch schedulers for cluster and grid infrastructures.

◮ Assert the value of information in such cases.



INDEX POLICIES FOR ROUTING

Optimal routing policy problem is still open for n different M/M/1Heuristic : index policy inspired from the Multi-Armed Bandit⇒ free parameter and compute an equilibrium point.[Mitrani 2005] for routing and repair problems.

µ

S servers

Capacity C

λ

W

b

µ

µ

W is the rejection cost (free parameter).

Theorem

There is an optimal policy of threshold type :there exists θ such that :Reject if x > θ and accept otherwise.- θ does not depend on C as long as C > θ (including if C is infinite).- θ is a non-decreasing function of W.



INDEX POLICIES FOR ROUTING(II)

Computation of θ(W) linear system of corresponding to Bellman’s equation, afteruniformization.

0

5

10

15

20

25

30

0 2 4 6 8 10 12 14 16 18 20

Bes

tThr

esho

ld(W

)

W

’TW.txt’

Index function I(x) = inf{W | θ(W) = x}.Indifference case : when queue size is x, rejecting or accepting the next batch are bothoptimal choices if the rejection cost is I(x).



SOME NUMERICAL EXPERIMENTS(I)

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

1-9_1-9_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

1.5

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

1-9_9-9.95_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

9-1_1-9_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

9-1_9.0-9.95_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Several cases with two queues with respective parameters (µ = 9, 1), C = 100



SOME NUMERICAL EXPERIMENTS(III)

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

1.18

1.2

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-2_1-9_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-2_9.0-9.95_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

1.18

1.2

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-2_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-2_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Several cases with two queues with respective parameters (µ = 8, 2), C = 100.



SOME NUMERICAL EXPERIMENTS(IV)

0.95

1

1.05

1.1

1.15

1.2

1.25

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8.1-1.9_1-9_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8.1-1.9_9.0-9.95_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.99

1

1.01

1.02

1.03

1.04

1.05

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

7-3_1-9_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.995

1

1.005

1.01

1.015

1.02

1.025

1.03

1.035

1.04

1.045

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

7-3_9.0-9.95_100.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Some other cases



SOME NUMERICAL EXPERIMENTS(V)

0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-1-1_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

8-1-1_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

1.3

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

7-2-1_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

7-2-1_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Three queues. Respectively, µ = 8, 1, 1 and µ = 7, 2, 1.



NUMERICAL EXPERIMENTS(VI)

0.95

1

1.05

1.1

1.15

1.2

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

6-3-1_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

6-3-1_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.96

0.98

1

1.02

1.04

1.06

1.08

1.1

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

6-2-2_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.97

0.98

0.99

1

1.01

1.02

1.03

1.04

1.05

1.06

1.07

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

6-2-2_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Now, µ = 6, 3, 1 and µ = 6, 2, 2.



NUMERICAL EXPERIMENTS(VII)

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

1.14

1.16

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

5-4-1_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.98

1

1.02

1.04

1.06

1.08

1.1

1.12

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

5-4-1_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.97

0.975

0.98

0.985

0.99

0.995

1

1.005

1.01

1.015

1.02

1.025

1 2 3 4 5 6 7 8 9

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

5-3-2_1-9_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.97

0.975

0.98

0.985

0.99

0.995

1

1.005

1.01

1.015

1.02

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Aver

age

wai

ting t

ime

(rat

io w

ith o

pti

m)

Load

5-3-2_9.0-9.95_50.txt

JSQJSQ-mu

JSQ-mu2Seq

*optim*

Now, µ = 5, 4, 1 and µ = 5, 3, 2.



ROBUSTNESS OF INDEX POLICIES

The index policy was computed for λ = 5 or 9 and used over the whole range λ = 1 to10.

0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1 2 3 4 5 6 7 8 9

Ave

rage

wai

ting

time

(rat

io w

ith o

ptim

)

Load

9-1_1-9_50.txt.1

mu : 9, 1c : 1, 1Nmax : 50 lambda index : 5

JSQJSQ-mu

JSQ-mu2Seq

*optim*

0.95

1

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10

Ave

rage

wai

ting

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1.25

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HTTP://PSI.GFORGE.INRIA.FR


http://psi.gforge.inria.fr


SIMULATION WORKFLOW

Samples

Coupling timeSample of rewards

Single trajectory

Steady-state independentIndependent trajectories R, S-plus,...

User defined scripts

Statistical analyzer

Simulation kernels

Forward samplingtrajectories

Backward samplingMonotone

Action of the event

Queues descriptionservercs, capacities

Event descriptionRateActivation condition

Model libraries

Samples

Coupling timeSample of rewards

Single trajectory

Steady-state independentIndependent trajectories R, S-plus,...

User defined scripts

Statistical analyzer



SIMULATION KERNELS

Trajectory Sampling

◮ Initial state

◮ Seed for the random scheme

=⇒ sequence of (time,state)

Set of trajectories

0

10

20

30

40

50

0 200 400 600 800 1000 1200 1400 1600

Traj0Traj1Traj2Traj3Traj4Traj5Traj6Traj7Traj8Traj9

Steady-state Sampling

◮ Monotone events, (envelopes)

◮ Seed for the random scheme

=⇒ sample of steady-state


Time

0

−i

−j

−τ ∗

Stationary Process

X

X

X

X

Zi

Zj

Z−τ∗ = {X0}

Z0 = X

collapse



PARALLEL SAMPLING

Computing ressources

◮ Many cores machine (desktop (4), server (16) or parallel (48))

◮ Parallel programing frameworks (OpenMP)

Objective

◮ Evaluate efficiency of parallel implementations on multicore platforms

Forward Simulation

◮ Space parallel approach

Generate a separate Markov chain on each core and combine results [Glynn92]

◮ Time parallel approach

Divide the iteration space and try to precompute parts of the Markov chain [Nicol94]

◮ Space-time parallel approach

Divide the model in several Markov chains [HsiehG09]

Backward Simulation

◮ Generate samples on each core (natural approach)



DEMONSTRATION

Philosophy

◮ C code

◮ Text files

◮ Models / Kernels / Results

◮ Model generation/ Statistical analysis is out of the software

A simple example

Lambda = 1.6

Capacity: 50

Queue 1 Queue 2

Capacity: 50

Mu = 1.8 Thêta = 2

OverflowOverflow

A more complex example

8

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

0

1

2

3

4

5

6

7

10

9



SYNTHESIS

Sampling of Poisson Systems

◮ Several approaches depending on the objective

◮ Common model description : events and distribution

Ψ3 Software

◮ Transient trajectories : Forward methods (Parallel, I/O pipelining, Nicol’s scheme)

◮ Steady-state sampling : Backward methods (Perfect monotone, envelopes)

Future work (practical)

◮ Tuning the sampling methods : adaptive

◮ Impact of I/O : in-situ analysis.

Future work (theoretical)

◮ Coupling time : partial coupling, hitting time,...

◮ Measure transformation : importance sampling, variance reduction

◮ Coupling time, mixing time, spectral gap, entropy

Reproducibility : Everything is on : http://psi.gforge.inria.fr


http://psi.gforge.inria.fr

poisson systems and perfect sampling -...

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